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The perpendicular form of the line 3xy+4=0  is

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a
xcos5π6+ysin5π6=2
b
xcosπ+ysinπ=3
c
xcosπysinπ=4
d
xcos3π2ysin3π2=3

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detailed solution

Correct option is A

3xy+4=03x+1y=4

Divide with 2 an both sides we get

3x2+12y=42x(32)+y(12)=2

Here cosα=32,sinα=12,p=2

Hence α  is in Q2α=5π6

The equation n of the line in perpendicular form is xcos5π6+ysin5π6=2

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